Thực hiện phép nhân \(\begin{array}{l} (x + y)\left( {x + \frac{1}{2}y} \right)\left( {1 - \frac{{xy}}{3}} \right) \end{array}\) ta được
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai\(\begin{array}{l} (x + y)\left( {x + \frac{1}{2}y} \right)\left( {1 - \frac{{xy}}{3}} \right)\\ = \left( {x(x + y) + \frac{1}{2}y(x + y)} \right)\left( {1 - \frac{{xy}}{3}} \right)\\ = \left( {{x^2} + xy + \frac{{xy}}{2} + \frac{{{y^2}}}{2}} \right)\left( {1 - \frac{{xy}}{3}} \right)\\ = \left( {{x^2} + xy + \frac{{xy}}{2} + \frac{{{y^2}}}{2}} \right) - \frac{{xy}}{3}\left( {{x^2} + xy + \frac{{xy}}{2} + \frac{{{y^2}}}{2}} \right)\\ = \left( {{x^2} + xy + \frac{{xy}}{2} + \frac{{{y^2}}}{2}} \right) - \left( {{x^2} \cdot \frac{{xy}}{3} + xy \cdot \frac{{xy}}{3} + \frac{{xy}}{2} \cdot \frac{{xy}}{3} + \frac{{{y^2}}}{2} \cdot \frac{{xy}}{3}} \right)\\ = \left( {{x^2} + xy + \frac{{xy}}{2} + \frac{{{y^2}}}{2}} \right) - \left( {\frac{{{x^3}y}}{3} + \frac{{{x^2}{y^2}}}{3} + \frac{{{x^2}{y^2}}}{6} + \frac{{x{y^3}}}{6}} \right)\\ = {x^2} + xy + \frac{{xy}}{2} + \frac{{{y^2}}}{2} - \frac{{{x^3}y}}{3} - \frac{{{x^2}{y^2}}}{3} - \frac{{{x^2}{y^2}}}{6} - \frac{{x{y^3}}}{6}\\ = {x^2} + \frac{{3xy}}{2} + \frac{{{y^2}}}{2} - \frac{{{x^3}y}}{3} - \frac{{{x^2}{y^2}}}{2} - \frac{{x{y^3}}}{6} \end{array}\)
